# Find Laplace Transform using unit step function given graph of a periodic impulse function. (5.3-33)

Please correct my work. The textbook answer which is expressed exactly like this $1/s(1+e^{-s})$ does not match my own.

Find the Laplace Transform for one period of the perpetual periodic function shown in graph.

First express function using the unit step:

$$f(t)= u(t)-u(t-1)$$

The Laplace Transform is easily found:

$$f(t)= \frac{1}{s}-\frac{e^{-s}}{s}= \frac{1-e^{-s}}{s}$$

Where did I go wrong?

Here is a very simple method to solve this problem. Note that $f(t)-f(t-2)$ is just a single period of $f(t)$ in the range $[0,2]$, i.e.

$$f(t)-f(t-2)=u(t)-u(t-1)\tag{1}$$

the Laplace transform of which you've already found:

$$\mathcal{L}\{u(t)-u(t-1)\}=\frac{1-e^{-s}}{s}\tag{2}$$

But you also have

$$\mathcal{L}\{f(t)-f(t-2)\}=F(s)-e^{-2s}F(s)=F(s)(1-e^{-2s})\tag{3}$$

where $F(s)$ is the desired Laplace transform of the periodic function $f(t)$. Equating (2) and (3) gives

$$F(s)=\frac{1-e^{-s}}{s(1-e^{-2s})}=\frac{1}{s(1+e^{-s})}$$

because

$$\frac{a-b}{a^2-b^2}=\frac{a-b}{(a+b)(a-b)}=\frac{1}{a+b}$$

Your $f(t)$ is correct for the imagine provided if you only take one period. $$f(t) = \mathcal{U}(t)-\mathcal{U}(t-1)$$ Now take the Laplace transform of $$\int_0^{\infty}f(t)e^{-st}dt = \int_0^{1}e^{-st}dt$$

If you want to find the Laplace transform of the periodic unit step function, then first we will write $f(t)$ as a series. $$f(t) = \sum_{n=0}^{\infty}\bigl[\mathcal{U}(t-2n)-\mathcal{U}[t-(2n+1)]\bigr]$$ Let's write out a few terms. \begin{align} f(t) &= \mathcal{U}(t)-\mathcal{U}(t-1)+\mathcal{U}(t-2)-\mathcal{U}(t-3)+\mathcal{U}(t-4)-\cdots\\ & = \mathcal{U}(t) + \sum_{n=1}^{\infty}(-1)^n\mathcal{U}(t-n)\\ \mathcal{L}\{f(t)\} &= \int_0^{\infty}e^{-st}dt + \int_0^{\infty}\sum_{n=1}^{\infty}(-1)^n\mathcal{U}(t-n)e^{-st}dt \end{align} We can interchange the $\sum(-1)^n$ with integral since $\mathcal{U}(t-n)e^{-st}\geq 0$ for $t \geq 0$. \begin{align} &= \frac{1}{s} + \sum_{n=1}^{\infty}(-1)^n\int_n^{\infty}e^{-st}dt\\ &= \frac{1}{s} + \sum_{n=1}^{\infty}(-1)^n\frac{e^{-sn}}{s}\\ &= \frac{1}{s}\Bigl[1 - e^{-s}\sum_{n=0}^{\infty}(-1)^ne^{-sn}\Bigr]\\ &= \frac{1}{s}\Bigl[1 - \frac{e^{-s}}{1+e^{-s}}\Bigr] \end{align}

• I assumed I only needed one period to solve it correctly but if that is not correct how should I go about solving for the entire function (to infinity)? At best my textbook does a very sub par explanation of the unit step function. – Jules Manson Feb 3 '15 at 3:00
• Thank you again. I had thought I may have to use a summation series but I was hoping for something simpler like an integral. – Jules Manson Feb 3 '15 at 3:06
• sorry please disregard this comment – Jules Manson Feb 3 '15 at 3:09
• I got the idea to use just one period (cycle) from this formula: $$ℒ\{f(t)\}=\frac{ℒ\{f_​1​​(t)\}}{1−e​^{−sp}}$$ where p is the period and $f_1$ is the function for one period. I tried it but I could not get the same answer as in text. I found it at intmath.com/laplace-transformation/… – Jules Manson Feb 3 '15 at 3:14